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Fix a nonnegative integer $n$. Let $P(x)$ be a polynomial of degree $n$ (over the real numbers) such that for all $k\in\left\{0,1,\ldots,n\right\}$, we have $P(k)=\dfrac{k}{k+1}$. Find $P(n+1)$.

My attempt:

I consider a new polynomial $Q(x)=(x+1)P(x)-x$. This $Q$ is a polynomial of degree $\leq n+1$ satisfying: $Q(k)=0\quad\forall k\in\{0,1,2,\ldots,n\}$. Therefore $Q(x)=\lambda x(x-1)\cdots(x-n)$ for a certain real $\lambda$. Hence, $Q(-1)=\lambda \cdot (-1)^{n+1}(n+1)!$, but also $Q(-1)=(-1+1)P(-1)+1=1$. Comparing these, we get $\lambda \cdot (-1)^{n+1} (n+1)! = 1$, thus $\lambda=\dfrac{(-1)^{n+1}}{(n+1)!}$. Since $Q(n+1)=\lambda(n+1)!$ then $Q(n+1)=(-1)^{n+1} $, and from $P(x) = \dfrac{Q(x)+x}{x+1}$ we deduce: $$P(n+1)=\dfrac{(-1)^{n+1}+n+1}{n+2}.$$ I doubted that the answer were $P(n+1)=\frac{n+1}{n+2}$ but the term $(-1)^{n+1}$ wasn’t expected. Is there any mistake here? Does someone have an alternative proof?

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  • $\begingroup$ Try a couple of examples for small $n$. This won't prove that you are right but it may improve your confidence. It might prove that you are wrong. $\endgroup$ – badjohn Apr 26 '19 at 16:15
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Your solution is fine.$$Q(-1) = \lambda (-1)(-2)\ldots (-(n+1))=\lambda (-1)^{n+1}(n+1)!=1$$

$$\lambda = \frac{(-1)^{n+1}}{(n+1)!}$$

$$Q(x)=\frac{(-1)^{n+1}x(x-1)\ldots (x-n)}{(n+1)!}=(x+1)P(x)-x$$

Let $x=n+1$

$$\frac{(-1)^{n+1}(n+1)n\ldots 1}{(n+1)!}=(n+2)P(n+1)-(n+1)$$

Hence $$P(n+1) = \frac{n+1 + (-1)^{n+1}}{n+2}$$

Let's check with small cases to see if $(-1)^{n+1}$ should be there.

Let $n=1$, then we have $P(0)=0$ and $P(1)=\frac12$. Then $P(x) = \frac{x}2$.

We have $P(2)= 1$.

Notice that $\frac{n+1}{n+2}<1$ and hence $P(n) = \frac{n+1}{n+2}$ is certainly wrong.

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  • $\begingroup$ That is the same as mine. Do you know an alternative proof avoiding the use of this specific polynomial $Q$? Typo fixed, forgot the frac. $\endgroup$ – DINEDINE Apr 26 '19 at 16:22
  • $\begingroup$ nope, your approach seems elegant enough. $\endgroup$ – Siong Thye Goh Apr 26 '19 at 16:24

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